Question 1179576
It appears you're dealing with a Chi-Square Goodness of Fit test. Here's how to approach the problem:

**1. State the Hypotheses**

* Null Hypothesis (H0): The distribution of disorders in 2018 is the same as in 2010.
* Alternative Hypothesis (H1): The distribution of disorders in 2018 is different from 2010.

**2. Calculate Expected Frequencies**

Multiply the percentages from 2010 by the sample size in 2018 (300) to get the expected counts for each disorder:

| Disorder | 2010 % | Expected Count |
|---|---|---|
| Schizophrenia | 54.9% | 164.7 |
| Major Depression | 21.1% | 63.3 |
| Obsessive-Compulsive | 7.9% | 23.7 |
| Anxiety | 4.5% | 13.5 |
| Personality | 2.9% | 8.7 |
| Other | 8.8% | 26.4 |

**3. Calculate the Chi-Square Test Statistic**

The formula for the Chi-Square test statistic is:

χ² = Σ [(Observed Count - Expected Count)² / Expected Count]

| Disorder | Observed | Expected | (O - E)² / E |
|---|---|---|---|
| Schizophrenia | 158 | 164.7 | 0.27 |
| Major Depression | 65 | 63.3 | 0.05 |
| Obsessive-Compulsive | 29 | 23.7 | 1.18 |
| Anxiety | 11 | 13.5 | 0.46 |
| Personality | 14 | 8.7 | 3.22 |
| Other | 23 | 26.4 | 0.44 |

χ² = 0.27 + 0.05 + 1.18 + 0.46 + 3.22 + 0.44 = **5.62**

**4. Determine Degrees of Freedom**

Degrees of freedom (df) = Number of categories - 1 = 6 - 1 = 5

**5. Find the P-value**

Using a Chi-Square distribution table or calculator with df = 5 and χ² = 5.62, you'll find the P-value.  The P-value is approximately 0.346.

**6. Compare P-value to Significance Level (α)**

* α = 0.01 (given)
* P-value (0.346) > α (0.01)

**7. Make a Decision**

Since the P-value is greater than α, we **Do Not Reject the Null Hypothesis**.

**8. Conclusion**

**There is not enough evidence to suggest that the breakdown of patients by disorder has changed significantly since 2010.**

**Important Note:** While the observed counts in the 2018 sample might look different from the 2010 percentages, the Chi-Square test tells us that this difference is not statistically significant. The variation could be due to random chance in the sample.